Modular Arithmetic: Multiplicative inverse

Question

I want to solve this question using mod arithmetic:

7/3 mod 8.

Somebody told me to do it using multiplicative inverse:

7 (Multiplicative inverse of 3) mod 8

I can't find any example related to the above method. Somebody please guide me.

Zulfi.

Please read this How to ask a good question to begin with and share your thoughts and efforts in the question — The Demonix _ Hermit, Dec 01 '19 at 16:37
I didn't solve the entire problem for you. I just showed you how to find a multiplicative inverse for $3\pmod 8$. — lulu, Dec 01 '19 at 16:48
So will the above question become: 7 * 9 mod 8, some body please guide me. — zak100, Dec 01 '19 at 16:54

score 0 · Accepted Answer · answered Dec 01 '19 at 16:39

0

Using multiplicative inverse would work, but also note that $7\equiv15\pmod 8$ — now can you solve your question using modular arithmetic?

answered Dec 01 '19 at 16:39

J. W. Tanner

First I want to solve the above one, then I would try this one. I can't understand the solution for my question. Why you put the "mod 8" in brackets? – zak100 Dec 01 '19 at 16:59
$3$ is its own inverse mod $8$, so $7/3\equiv 7\times3\pmod8; $ that gives the same answer as mine. Putting mod $8$ in parentheses is standard mathematical notation – J. W. Tanner Dec 01 '19 at 17:03
It would become "21 mod 8", that would be 5, is this correct, sorry your answer was 1. – zak100 Dec 01 '19 at 17:16
1

$5$ is correct; my answer was also $15/3=5$ – J. W. Tanner Dec 01 '19 at 17:17
I have two more questions: 9/6 mod 11, that would be 9 * (MI of 6) mod 11; 9 * 6 mod 11; 54 mod 11 i.e. 9 or it should be 3/2 mod 11, that would be 3 * (MI of 2) mod 11; 6 mod 11= 6, which answer is correct? Should I have to simplify first? Somebody please guide me. – zak100 Dec 01 '19 at 17:24
probably answered elsewhere on this site, but $9/6\equiv 9\times6^{-1}\equiv 9\times 2=18\equiv7\pmod{11}$ and also $9/6\equiv3/2\equiv3\times6=18\equiv7\pmod{11}$ – J. W. Tanner Dec 01 '19 at 17:26

1 Answers1